A basket ball sits in the ball cage in the gym motionless

How do you think the switch controls the flow of current to the light?

nadezda [96]

Answer:

A switch interrupts the flow of current in a circuit.

Explanation:

3 0

1 year ago

Two positive point charges are 4.9cm apart. If the electric potential energy is 70.0 μJ, what is the magnitude of the force betw

ExtremeBDS [4]

Hi there!

Recall the following:

$V \text{ (Electric Potential Energy) } = \frac{kq_1q_2}{r}\\\\F_E = \frac{kq_1q_2}{r^2}$

k = Coulomb's Constant (Jm/C²)

q = Charge (C)
r = distance between charges (m)

To calculate the electric force between the two charges, we can simply divide by another 'r' (distance):

$F_E = \frac{70}{0.049} = \boxed{1428.57 \mu J}$

6 0

2 years ago

A ball thrown vertically upward is caught by the thrower after 2.00 s. Find (a) the initial velocity of the ball and (b) the max

ankoles [38]

Answer:

a) 9.8 m/s

b) 4.9 m

Explanation:

This problem is a good example of Vertical motion, where the main equations for this situation are:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

$V^{2}={V_{o}}^{2}-2gy$ (2)

Where:

$y$ is the height of the ball at a given time

$y_{o}=0m$ is the initial height of the ball (assuming the hand of the thrower the origin of the system)

$V_{o}$ is the initial velocity of the ball

$V$ is the final velocity of the ball

$t=2s$ is the time it takes for the ball to make the complete movement (from the moment it is thrown until it falls back into the pitcher's hands)

$g=9.8 m/s^{2}$ is the acceleration due to gravity

Knowing this, let's begin with the answers:

<h3>a) Initial velocity </h3>

In order to find the initial velocity $V_{o}$ of the ball, we will use equation (1) and $t=2s$ , taking into account that $y=0 m$ and $y_{o}=0m$ at this given time:

$0=0+V_{o}t-\frac{1}{2}gt^{2}$ (3)

Isolating $V_{o}$ :

$V_{o}=\frac{1}{2}gt$ (4)

$V_{o}=\frac{1}{2}(9.8 m/s^{2})(2 s)$ (5)

Then:

$V_{o}=9.8 m/s$ (6)

<h3>b) Maximum height </h3>

In this part, we will use equation (2), knowing the value of the height is maximum when $V=0$ . So, we will name this height as $y_{max}$ :

$0={V_{o}}^{2}-2gy_{max}$ (7)

Isolating $y_{max}$ :

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8)

$y_{max}=\frac{{(9.8 m/s)}^{2}}{2(9.8 m/s^{2})}$ (9)

Finally:

$y_{max}=4.9 m$

4 0

3 years ago

4 The temperature at the center of the sun is 1.55 x 10' K. Express this in degrees

Sloan [31]

Answer:

The answer is D

Explanation:

3 0

3 years ago

A student in gym class swings from a rope and they are moving 5 m/s at the bottom of their swing. What is the height they reach

vaieri [72.5K]

Answer:

A

Explanation:

4 0

2 years ago